Simplify the following expression and state the condition under which the simplification is valid. You can assume that $k \neq 0$. $a = \dfrac{5k - 10}{6k} \times \dfrac{7k}{10k - 20} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $a = \dfrac{ (5k - 10) \times 7k } { 6k \times (10k - 20) } $ $ a = \dfrac {7k \times 5(k - 2)} {6k \times 10(k - 2)} $ $ a = \dfrac{35k(k - 2)}{60k(k - 2)} $ We can cancel the $k - 2$ so long as $k - 2 \neq 0$ Therefore $k \neq 2$ $a = \dfrac{35k \cancel{(k - 2})}{60k \cancel{(k - 2)}} = \dfrac{35k}{60k} = \dfrac{7}{12} $